I'm doing eigen values and eigen functions so the left expression has to equal the right.

myininaya

5 years ago

You said you wrote it wrong? Did I write it right?

anonymous

5 years ago

Sorry I deleted the wrong one already. The one I wrote is right. Anyways I should rephrase my questions. I want the left side to equal the right side. How can I change the left side in order to make the equation true so that the left hand side equals the right hand side?

myininaya

5 years ago

\[\frac{1-\sin^2(x)-\sin^2(x)}{\sin(x)}-2=\frac{1-2\sin^2(x)}{\sin(x)}-2=\frac{1-2\sin^2(x)-2\sin(x)}{\sin(x)}\]
hmmm.... I don't think they are the same ....
Did you see if you could find a counterexample?

myininaya

5 years ago

\[x=\frac{\pi}{2}\]
\[\frac{\cos^2(x)-\sin^2(x)}{\sin(x)}-2=\sin^2(x)\]
\[\frac{\cos^2(\frac{\pi}{2})-\sin^2(\frac{\pi}{2})}{\sin(\frac{\pi}{2})}=\frac{0-1}{1}=-1\]
But the other side is 1 when x=pi/2
Therefore both sides are not the same

myininaya

5 years ago

-1-2 does not equal 1

myininaya

5 years ago

no questions?

anonymous

5 years ago

What do you suggest I do in order to get them equal?

myininaya

5 years ago

they are not equal

myininaya

5 years ago

this is not an identity
I gave you a counterexample above
try pluggin' in pi/2 for x
we do not get the same thing on both sides
unless you meant to type something else